A recent study stated that if a person chewed gum the average number of sticks of gum he or she chewed daily was eight. To teat the claim a researcher selected a random sample of 36 gum chewers and found the mean number of sticks of gum chewed per day was 9 and the standard deviation was 1. At alpha equals 0.05 is the number of sticks of gum a person chews per day actually greater than 8.

I'll give you a few hints to help get you started on this one.

Ho: µ = 8 --->meaning: the population mean is equal to 8.
Ha: µ > 8 --->meaning: the population mean is greater than 8.

Using a one-sample z-test formula to find the test statistic:
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
z = (9 - 8)/(1/√36)

Finish the calculation and determine whether or not to reject the null hypothesis (Ho). If the null is rejected, then µ > 8.

I hope this will help.

6.16

To test if the number of sticks of gum a person chews per day is actually greater than 8, we can use a one-sample t-test.

Here are the steps to conduct the hypothesis test:

Step 1: State the null and alternative hypotheses:

Null hypothesis (H0): The average number of sticks of gum chewed per day is equal to 8.
Alternative hypothesis (Ha): The average number of sticks of gum chewed per day is greater than 8.

Step 2: Determine the significance level (alpha):

The significance level (alpha) is given in the problem as 0.05, which means we will reject the null hypothesis if the p-value is less than 0.05.

Step 3: Calculate the test statistic:

The test statistic for a one-sample t-test is calculated using the formula:

t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))

In this case:
Sample mean (x̄) = 9
Hypothesized mean (μ) = 8
Sample standard deviation (s) = 1
Sample size (n) = 36

t = (9 - 8) / (1 / sqrt(36)) = 3

Step 4: Determine the rejection region:

Since the alternative hypothesis is one-sided (greater than), we will look for the critical value from the right-tail of the t-distribution. Since df = n - 1 = 36 - 1 = 35, we can find the critical value using a t-table or calculator.

At alpha = 0.05 with 35 degrees of freedom, the critical t-value is approximately 1.689.

Step 5: Compare the test statistic with the critical t-value:

Since the test statistic (t = 3) is greater than the critical t-value (1.689), it falls in the rejection region.

Step 6: Calculate the p-value:

The p-value is the probability of obtaining a test statistic as extreme as the observed, assuming the null hypothesis is true.

Using a t-distribution table or calculator, we can find that the p-value for t = 3 with 35 degrees of freedom is less than 0.001.

Step 7: Make a decision:

Since the p-value (less than 0.001) is less than the significance level (0.05), we reject the null hypothesis.

Step 8: Draw a conclusion:

Based on the sample data, there is sufficient evidence to conclude that the number of sticks of gum a person chews per day is actually greater than 8.

Note: It is important to be cautious when generalizing the conclusion to the entire population, as the study only considered a random sample of 36 gum chewers.

To test the claim and determine if the number of sticks of gum a person chews per day is actually greater than 8, we can perform a hypothesis test using the given data.

Step 1: Formulate the null hypothesis (H₀) and the alternative hypothesis (H₁):

Null hypothesis (H₀): The average number of sticks of gum chewed per day is equal to 8.
Alternative hypothesis (H₁): The average number of sticks of gum chewed per day is greater than 8.

Step 2: Determine the significance level (alpha) and the test statistic:

Given alpha (α) = 0.05 (5% level of significance).

Test statistic: We'll use a t-test since the sample size is relatively small (n < 30) and the population standard deviation is unknown.

Step 3: Calculate the test statistic:

The test statistic (t-score) can be calculated using the following formula:
t = (sample mean - hypothesized mean) / (standard deviation / √sample size)

Given:
Sample mean (x̄) = 9
Hypothesized mean (μ₀) = 8
Standard deviation (σ) = 1
Sample size (n) = 36

t = (9 - 8) / (1 / √36)
= 1 / (1/6)
= 1 * 6
= 6

Step 4: Determine the critical value:

Since we have a one-tailed test (looking for a greater than), we need to find the critical value corresponding to the given significance level (α) and degrees of freedom (n - 1 = 36 - 1 = 35) in a t-distribution table.

At α = 0.05 with 35 degrees of freedom, the critical value is approximately 1.69.

Step 5: Compare the test statistic with the critical value and make a decision:

Since the test statistic (t = 6) is greater than the critical value (1.69), we have enough evidence to reject the null hypothesis.

Step 6: State the conclusion:

Based on the sample data, at a significance level of 0.05, we have sufficient evidence to conclude that the number of sticks of gum a person chews per day is actually greater than 8.

Please note that this conclusion is specific to the given sample and cannot be generalized to the entire population of gum chewers.